Problem 28 - Calculus Early Transcendentals

Question

Find \(F^{\prime}(\pi)\) given that \(f(\pi)=10, f^{\prime}(\pi)=-1, g(\pi)=-3\) and \(g^{\prime}(\pi)=2.\) (a) \(F(x)=6 f(x)-5 g(x)\) (b) \(F(x)=x(f(x)+g(x))\) (c) \(F(x)=2 f(x) g(x)\) (d) \(\quad F(x)=\frac{f(x)}{4+g(x)}\)

Step-by-Step Solution

Answer

(a) -16, (b) \(7 + \pi\), (c) 46, (d) -21

Step 1: Understand the Problem

We need to find the derivative of each function, denoted as \( F^{\prime}(x) \), and evaluate it at \( x = \pi \). We will use the product rule, quotient rule, and basic derivative rules to find these derivatives.

Step 2: Differentiate Part (a)

The function is \( F(x) = 6f(x) - 5g(x) \). The derivative is obtained by differentiating each term: \( F^{\prime}(x) = 6f^{\prime}(x) - 5g^{\prime}(x) \). Substituting \( x = \pi \), we get \( F^{\prime}(\pi) = 6(-1) - 5(2) = -6 - 10 = -16 \).

Step 3: Differentiate Part (b)

The function is \( F(x) = x(f(x) + g(x)) \). Using the product rule, \( F^{\prime}(x) = 1\cdot(f(x) + g(x)) + x(f^{\prime}(x) + g^{\prime}(x)) \). Substituting the values at \( x = \pi \), we get \( F^{\prime}(\pi) = (10 - 3) + \pi((-1) + 2) = 7 + \pi \).

Step 4: Differentiate Part (c)

The function is \( F(x) = 2f(x)g(x) \). Using the product rule, \( F^{\prime}(x) = 2(f^{\prime}(x)g(x) + f(x)g^{\prime}(x)) \). Substituting \( x = \pi \), we find \( F^{\prime}(\pi) = 2((-1)(-3) + 10(2)) = 2(3 + 20) = 46 \).

Step 5: Differentiate Part (d)

The function is \( F(x) = \frac{f(x)}{4+g(x)} \). Using the quotient rule, \( F^{\prime}(x) = \frac{f^{\prime}(x)(4 + g(x)) - f(x)g^{\prime}(x)}{(4 + g(x))^2} \). Substituting \( x = \pi \), we have \( F^{\prime}(\pi) = \frac{(-1)(4 - 3) - 10(2)}{(4 - 3)^2} = \frac{-1 - 20}{1} = -21 \).

Key Concepts

Product Rule Mastery

The product rule is a fundamental concept in calculus that is essential for differentiating products of two or more functions. When faced with a function like the one described in part (b) and (c), knowing the product rule can make the task much simpler.
The product rule states: if we have two functions, say, \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x)v(x) \) is given by: \[ (uv)^{ ext{'}} = u^{ ext{'}}v + uv^{ ext{'}} \] This rule allows us to see that the derivative of the product is not simply the product of the derivatives, as one might mistakenly think.
In our provided exercise, for \( F(x) = x(f(x) + g(x)) \), applying the product rule means treating \( x \) as one function and \( (f(x) + g(x)) \) as another. Ensuring accurate application involves differentiating \( x \) (which gives us 1, since the derivative of \( x \) is 1) and then \( (f(x) + g(x)) \) using simple derivative rules before substituting back into our product rule formula.
This produces: - \( F^{ ext{'}}(x) = 1 \cdot (f(x)+g(x)) + x(f^{ ext{'}}(x)+g^{ ext{'}}(x)) \) showing a clear practical application of the rule.

Quotient Rule Insights

When you need to differentiate a function that is the quotient of two other functions, the quotient rule becomes vital. Part (d) from our example exercise is a perfect candidate for this rule.
The quotient rule states that for two functions \( u(x) \) (numerator) and \( v(x) \) (denominator), the derivative of their quotient \( \frac{u(x)}{v(x)} \) is: \[ \left(\frac{u}{v}\right)^{\text{'}} = \frac{u^{\text{'}}v - uv^{\text{'}}}{v^2} \] It is important to remember that the order and signs are crucial here, differing from the product rule.
In the problem, this rule was applied to \( F(x) = \frac{f(x)}{4+g(x)} \). Here: - \( u(x) = f(x) \) and \( v(x) = 4 + g(x) \), hence \( v^{2} = (4+g(x))^2 \).
Substituting these into the quotient rule formula showed us how to manage the negative interactions often present in such problems. The calculation was simplified by substituting the values at \( x = \pi \). Mastering these subtleties can bolster your calculus prowess immensely.

Calculus Concepts Explained

Calculus, often seen as the mathematical study of change, offers powerful tools for analyzing complex systems through differentiation and integration.
At its core, the field is split into two branches:

• Differential calculus, concerned with rates of change and slopes of curves,
• Integral calculus, concerned with accumulation of quantities and the areas under and between curves.

In our exercise, differentiation plays a crucial role in finding the instantaneous rate of change of a function at a given point \( x = \pi \).
Understanding how to apply rules like the product rule and the quotient rule is key to solving various derivative problems. Whether it's finding the slope of a tangent line to a curve at a specific point or determining how a system changes over time, these concepts serve as vital tools. Delving deeper into calculus will uncover more advanced techniques, but a firm grasp of the differential basics, as shown in the example, lays a solid groundwork for further exploration.